Limits Question 3
Question 3 - 29 January - Shift 1
Let $x=2$ be a root of the equation $x^{2}+p x+q=0$
and $f(x)= \Bigg \lbrace \begin{matrix} \frac{1-\cos (x^{2}-4 p x+q^{2}+8 q+16)}{(x-2 p)^{4}}, & x \neq 2 p \\ 0 & x=2 p\end{matrix} . $
Then $\lim _{x \to 2^{+}}[f(x)]$
where [ . ] denotes greatest integer function, is
2
1
0
-1
Show Answer
Answer: (3)
Solution:
Formula: L’Hospital’s rule, Algebra of limits
$ \begin{aligned} & \lim _{x \to 2 p^{+}}(\frac{1-\cos (x^{2}-4 p x+q^{2}+8 q+16)}{(x^{2}-4 p x+q^{2}+8 q+16)^{2}})(\frac{(x^{2}-4 p x+q^{2}+8 q+16)^{2}}{(x-2 p)^{2}}) \\ & \lim _{h \to 0} \frac{1}{2}\left(\frac{(2 p+h)^{2}-4 p(2 p+h)+q^{2}+82+16}{h^{2}}\right)^{2}=\frac{1}{2} \end{aligned} $
Using L’Hôpital’s
$ \lim _{x \to 2 p^{+}}[f(x)]=0 $
BETA
